Consider the figure eight curve $\gamma : (-\pi, \pi) \to \mathbb{R}^2$ defined by $\gamma(t) = (\sin(2t), \sin(t))$. Show that the velocity vector field of $\gamma$ is not extensible.
The notion of extensible vector fields along curves is a bit cryptic. I have one condition for a velocity vector field of a path $\gamma$ to not be extensible and it is that if $\gamma(t_1)=\gamma(t_2)$, but $\gamma'(t_1) \ne \gamma'(t_2)$, then $\gamma'$ is not extensible. In this case $\gamma$ is injective so $\gamma(t_1)=\gamma(t_2) \implies t_1=t_2$ and so $\gamma'(t_1)=\gamma'(t_2)$ is a tautology.
The concrete definition for extensibility is that if $\gamma :I \to M$ is a smooth curve and $V$ a vector field along $\gamma$ there exists $\widetilde{V}$ a smooth vector field on $U \subset M$ containing the image of $\gamma$ such that $\widetilde{V}$ restricted to $\gamma$ gives $V$.
Now I'm quite sure that the origin gives some problems here, but I don't know how to make this into an argument?
Best Answer
The condition that you wrote down for the velocity velocity field of the path to be non-extensible is not far off.
Intuitively, we would love to apply this condition at $t_1 = 0$ and $t_2 = \pi$. We would say that $\gamma(t_1) = \gamma(t_2) = (0, 0)$, but $\gamma'(t_1) = (2, 1)$ and $\gamma'(t_2) = (2, -1)$, and then we would apply your condition to conclude that $\gamma$ cannot possibly be extensible. But sadly, this reasoning is invalid because your curve $\gamma$ is not actually defined at $t = t_2$.
However, we can strengthen your condition, as follows:
Proof of revised condition: Suppose that $U \subset \mathbb R^n$ is an open set containing the image of $\gamma$, and suppose that $\widetilde V$ is a smooth vector field on $U$ such that $\widetilde V|_{\gamma(t)} = \gamma'(t)$ at all $t \in I$. Since $\lim_{n \to \infty} \gamma(t_n) = \gamma(t_\star)$, we must have $\lim_{n \to \infty} \widetilde V|_{\gamma(t_n)} = \widetilde V|_{\gamma(t_\star)},$ by the continuity of $\widetilde V$. But $\lim_{n \to \infty} \widetilde V|_{\gamma(t_n)} = \lim_{n \to \infty} \gamma'(t_n)$ and $ \widetilde V|_{\gamma(t_\star)} = \gamma'(t_\star)$. This contradicts the fact that $\lim_{n \to \infty} \gamma'(t_n) \neq \gamma'(t_\star)$.
Now let's apply this to your problem. Take $t_n = \pi \left( 1 - \frac 1 n \right)$, and take $t_\star = 0$. We have $$\lim_{n \to \infty} \gamma(t_n) = \gamma(t_\star) = (0, 0).$$ However, we have $$ \lim_{n \to \infty} \gamma'(t_n) = (2, -1) \neq (2, 1) = \gamma'(t_\star).$$ Applying our revised condition, we see that $\gamma'$ is not extensible.